Oracle Regex 从字符串中的键值对中获取值

Question

Consider the following column row:

col
-------------------------
'{"day":"8","every":"2"}'

I am trying to get 8 from this string using regular expression to figure out the day.

so far I have:

SELECT 
    regexp_replace(col, '{"day":[^0-9]', '') as "day"
FROM 
   mytable;

This gives me:

 day
 ---------------
 8","every":"2"}

I am having trouble figuring out how to filter out the rest of the string from the first number forward. In my example I just want the number 8 for this row.

Answer 1

When you are lucky enough to use Oracle 12c Release 1 (12.1.0.2) or later, do take a look at JSON_VALUE

WITH t (s)
AS (
    SELECT '{"day":"8","every":"2"}'
    FROM DUAL
    )
SELECT JSON_VALUE(s, '$.day'  ) AS day
    ,  JSON_VALUE(s, '$.every') AS every
FROM t;

DAY   EVERY
---   -----
8     2

Answer 2

How about this?

SELECT 
    regexp_replace(col, '{"day":"([0-9]+).*', '\1') as "day"
FROM 
   mytable;

Answer 3

If you don't have access to JSON_VALUE() then I would recommend the following regex unless you always know the position of the day key in the JSON string:

SELECT REGEXP_REPLACE(col, '^.*"day":"(\d+)".*$', '\1') AS day
  FROM mytable;

This will replace the entire string (assuming it matches!) with the contents of the first capturing group (enclosed in parentheses: (\\d+) ). \\d indicates a digit 0-9 . If you want to return NULL values as well, you can replace \\d+ with \\d* . If negative or non-numeric values are possible, then I would recommend the following:

SELECT REGEXP_REPLACE(col, '^.*"day":"([^"]*)".*$', '\1') AS day
  FROM mytable;

This will return whatever characters that might be contained in the day key.

FYI, once you have the value, numeric or non-, you can convert it to a number safely by using TO_NUMBER() along with REGEXP_SUBSTR() :

SELECT COALESCE( TO_NUMBER( REGEXP_SUBSTR( REGEXP_REPLACE( col, '^.*"day":"[^"]*".*$', '\1' ), '\d+' ) ), 0 ) AS day
  FROM mytable;

Hope it helps.